{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Solution%20Assn.%202

Solution%20Assn.%202 - Copyright UC Regents Davis campus...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Copyright © UC Regents Davis campus, 2005-11. All Rights Reserved. May not be re-distributed without prior written consent of course instructor. Solutions to Assignment 2 Note: This problem is similar to Problem 1-20 on Page 28 of your Segel book. Problem 25 The ionization of H 3 PO 4 is shown below: H + + H 2 PO 4 - H + + HPO 4 = H + + PO 4 H 3 PO 4 Part a) In this part of the problem you will use the K a equation as follows to determine the hydrogen ion concentration and then calculate the pH from that.: [H + ] [H 2 PO 4 - ] As you see from the equation, for each H there is a H2PO4. Thus you can let each = x. K a = [H 3 PO 4 ] Look up the first pK a for phosphoric acid in Appendix IV of the Segel book. (It is 2.12) and then take the antilog of -2.12 to get the Ka, which is 7.59 x 10 -3 M. (Since the concentration of the acid is given as .2 M and it is less than 100 x the K a , you will have to include the removal of x in the denominator of the equation.) K a = 7.59 x 10 -3 M = x 2 0.2M - x x 2 = 1.52 x 10 -2 - 7.59 x 10 -3 x x 2 + 7.59 x 10 -3 x – 1.59 x !0 -2 = 0 Use the quadratic equation to solve for x. x = 34.94 x 10 -3 M = [H + ] pH = - log [H + ] = - log .03494 M = 1.457
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern